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We are dealing with $z \in \mathbb{C}$. I know that $$ \left(1+ \frac{z}{n} \right)^n \to e^{z} $$ as $n \to \infty$. So intuitively if $z_n \to z$ then we should have $$ \left(1+ \frac{z_n}{n} \right)^n \to e^{z}. $$ If $z_n \in \mathbb{R}$ I would be happy writing $$ \exp \left(n \log\left(1+ \frac{z_n}{n} \right) \right) = \exp \left(z_n \frac{\log\left(1+ \frac{z_n}{n} \right)-\log(1+0)}{\frac{z_n}{n}-0} \right) \to \exp(z \cdot 1) $$ where here we are using the definition of derivative. But if $z \in \mathbb{C}$ the logarithm is multivalued and I'm not sure that the same calculation is allowed. Is there a different way to show this for complex $z$?

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You do not need the logarithm function at all.

We begin with the bound, valid for complex $z$ with $|z|\leq 1$: $$|(1+z)-\exp(z)|\leq \left|{z^2\over2!}+{z^3\over 3!}+\cdots\right|\leq {|z|^2\over 2!}+{|z|^3\over 3!}+\cdots\leq |z|^2.$$ Similarly, we also have $|1+z|\leq \exp(|z|)$ and $|\exp(z)|\leq \exp(|z|)$ for all $z$.

Now suppose that $c_n\to c$ in the complex plane. Consider the telescoping sum $$w_1\cdots w_n-z_1\cdots z_n=\sum_{j=1}^n w_1\cdots w_{j-1}(w_j-z_j)z_{j+1}\cdots z_n,$$ and plug in $w_j=(1+c_n/n)$ and $z_j=\exp(c_n/n)$ to obtain $$\left(1+{c_n\over n}\right)^n-\exp(c_n)= \sum_{j=1}^n \left(1+{c_n\over n}\right)^{j-1}\left[\left(1+{c_n\over n}\right)-\exp(c_n/n)\right]\exp(c_n/n)^{n-j}.$$ For $n$ so large that $|c_n/n|\leq 1$, the bounds above give $$\left|\left(1+{c_n\over n}\right)^n-\exp(c_n)\right|\leq n \exp(|c_n|)\, {|c_n|^2\over n^2}\to 0\mbox{ as }n\to\infty.$$

This shows that $\left(1+{c_n\over n}\right)^n\to\exp(c)$ as $n\to\infty.$

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  • Your argument is fine, because $1+z_n/n\to 1$ necessarily, and so you only need $\log$ to be well behaved locally. Picking a consistent branch works fine.
  • On the other hand, you could 'squeeze' the result, by noting that the modulus difference from $e^z$ decays as $z_n\to z$. This argument works by continuity. (Edit: Fixed from only working for the real case, d'uh!)
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  • $\begingroup$ What does bounded above and below mean for complex numbers? $\endgroup$ – nullUser Apr 28 '13 at 0:39
  • $\begingroup$ Realized that as soon as I'd posted (: $\endgroup$ – Sharkos Apr 28 '13 at 0:43
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For your argument to be valid, specify a branch of $\log$; You can use the principal branch, $\text{Log}(z)$, but any branch that is analytic at $z=1$ should do:

$$\lim_{n\to 0}\left( 1 + \frac{z_n}{n}\right)^n = \lim_{n \to 0} \exp\left( n \text{ Log}\left( 1 + \frac{z_n}{n}\right)\right) = \cdots$$

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Brute force works too:

Suppose $|z| \le B$ for all $n$. Then \begin{eqnarray} | (1+{z \over n})^n - e^{z} |&=& |\sum_{k=0}^n {n! \over (n-k)! k!}{z^k \over n^k} - \sum_{k=0}^\infty {z^k \over k!}| \\ &=& |\sum_{k=2}^n \left( {n(n-1)\cdots (n-k+1) \over n^k} -1 \right) {z^k \over k!} | + | \sum_{k=n+1}^\infty {z^k \over k!} |\\ &\le& \sum_{k=2}^n \left| {n(n-1)\cdots (n-k+1) \over n^k} -1 \right| {B^k \over k!} + \sum_{k=n+1}^\infty {B^k \over k!} \end{eqnarray} Note that $\left| {n(n-1)\cdots (n-k+1) \over n^k} -1 \right| \le 2$.

Choose $\epsilon>0$. We can choose $N$ such that if $n \ge N$ then $\sum_{k=n+1}^\infty {B^k \over k!} < { 1\over 4} \epsilon$. Then \begin{eqnarray} | (1+{z \over n})^n - e^{z} | &\le& \sum_{k=2}^n \left| {n(n-1)\cdots (n-k+1) \over n^k} -1 \right| {B^k \over k!} + \sum_{k=n+1}^\infty {B^k \over k!} \\ &<& e^B \sum_{k=2}^N \left| {n(n-1)\cdots (n-k+1) \over n^k} -1 \right| + 2\sum_{k=N+1}^\infty {B^k \over k!} + { 1\over 4} \epsilon \\ &\le& e^B \sum_{k=2}^N \left| {n(n-1)\cdots (n-k+1) \over n^k} -1 \right| + { 3\over 4} \epsilon \end{eqnarray} Now choose $N' \ge N$ large enough so that $\left| {n(n-1)\cdots (n-k+1) \over n^k} -1 \right| < {1 \over N e^B} { 1\over 4} \epsilon$ whenever $n \ge N'$.

Then, if $n \ge N'$ we have $| (1+{z \over n})^n - e^{z} | < \epsilon$.

Now suppose $z_n \to z$. Then we have some $B$ such that $|z_n| \le B$ for all $B$. Let $\epsilon>0$, $N'$ be as above, and suppose $n \ge N'$.

Then $| (1+{z_n \over n})^n - e^{z} | \le | (1+{z_n \over n})^n - e^{z_n} | + |e^z - e^{z_n}| < \epsilon + |e^z - e^{z_n}|$.

Hence we see that $(1+{z_n \over n})^n \to e^{z}$.

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Simple proof by cheap tricks:

Say $f_n(z)=(1+z/n)^n$. Since $1+t\le e^t$ for $t\ge0$ it follows that $$|f_n(z)|\le e^{|z|}$$for all $z$.

So $(f_n)$ is a normal family. And $f_n(t)\to e^t$ for $t>0$. So a standard argument about normal families (someone's theorem; Vitali?) shows that $f_n\to\exp$ uniformly on compact sets. Hence if $z_n\to z$ then $f_n(z_n)\to\exp(z)$.


Standard Argument: Since $(f_n)$ has compact closure in the metric space $H(\Bbb C)$, if $f_n$ does not tend to $\exp$ in this metric space (i.e. uniformly on compact sets) there is a subsequence tending to something else. But that something else agrees with $\exp$ on the positive real axis, hence it's not something else after all.

In more detail:

Someone's Theorem Suppose $V$ is an open connected subset of the plane and $S\subset V$ has a limit point in $V$. Suppose $(f_n)\subset H(V)$ is a normal family, $f\in H(V)$, and $f_n(z)\to f(z)$ for all $z\in S$. Then $f_n\to f$ uniformly on compact subsets of $V$.

Proof. Let $||g||_K=\sup_{z\in K}|g(z)|$. Suppose $f_n$ does not tend to $f$ uniformly on compact sets. Then there exist a compact set $K$, a number $\epsilon>0$, and an infinite set $D\subset\Bbb N$ so that $$||f_n-f||_K\ge\epsilon\quad(n\in D).$$

Since $(f_n)$ is a normal family there is a sequence $(n_j)\subset D$ with $f_{n_j}\to g\in H(V)$ uniformly on compact sets. Uniform convergence on $K$ shows that $||g-f||_K\ge\epsilon$. But $g=f$ on $S$, hence $g=f$, contradiction. QED.

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You can use the following standard lemma:

Lemma: If $a_{n} $ is a sequence of complex numbers such that $n(a_{n} - 1)\to 0$ then $a_{n} ^{n} \to 1$.

Now let's put $$a_{n} =\dfrac{1+\dfrac{z_{n}}{n}}{1+\dfrac{z}{n}}$$ so that $$n(a_{n} - 1)=\frac{z_{n}-z}{1+z/n}\to 0$$ and then $a_{n} ^{n} \to 1$ and this means that $(1+(z_{n}/n))^{n}$ tends to the same limit as that of $(1+(z/n))^{n}$ and we are done.


The lemma mentioned at the beginning is proved easily by writing $a_{n} =1+b_{n}$ so that $nb_{n} \to 0$ and we have via binomial theorem $$|a_{n} ^{n} - 1|=|(1+b_{n})^{n}-1|=\left|nb_{n}+\frac{n(n-1)}{2!}b_{n}^{2}+\dots\right|$$ and clearly the RHS is bounded above by $$|nb_{n} |+|nb_{n} |^{2}+\cdots = \frac{|nb_{n} |} {1-|nb_{n}|}$$ Since $nb_{n} \to 0$ we are done.

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